Question

If $x=2+\sqrt{5}$ then find the value of $x^2+\frac{1}{x^2}$.

Answer 1

$\begin{array}{c}x=2+\sqrt{5}\\ x^2={\left(2+\sqrt{5}\right)}^2\\ =2^2+{\left(\sqrt{5}\right)}^2+2\left(2\right)\left(\sqrt{5}\right)\\ =4+5+5\sqrt{5}\\ =9+5\sqrt{5}\\ x^2+\frac{1}{x^2}=9+5\sqrt{5}+\frac{1}{9+5\sqrt{5}}\\ =\frac{{\left(9+5\sqrt{5}\right)}^2+1}{9+5\sqrt{5}}\\ =\frac{81+125+90\sqrt{5}+1}{9+5\sqrt{5}}\\ =\frac{207+90\sqrt{5}}{9+5\sqrt{5}}\end{array}$